BZOJ 5496: [2019省队联测]字符串问题 (后缀数组+主席树优化建图+拓扑排序)

题意

略

分析

考场上写了暴力建图40分溜了…(结果只得了30分)

然后只要优化建边就行了

首先给出的支配关系无法优化,就直接A向它支配的B连边.

考虑B向以B作为前缀的所有A连边,做一遍后缀数组,两个区间的左端点位置的LCP一定大于等于B的长度.在排好序的后缀数组上就是一段区间.然后用线段树优化建边.

那么用ST表倍增算一下能拓展的区间就行了.

然后由于只能向长度大于自己的连边,就按长度排序然后线段树改成主席树.

写+调一个下午…发现是sb错误…

CODE

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
char cb[1<<15],*cs=cb,*ct=cb;
#define getc() (cs==ct&&(ct=(cs=cb)+fread(cb,1,1<<15,stdin),cs==ct)?0:*cs++)
template<class T>inline void read(T &res) {
    char ch; while(!isdigit(ch=getc()));
    for(res=ch-'0';isdigit(ch=getc());res=res*10+ch-'0');
}
const int MAXS = 200005;
const int MAXA = 200005;
const int MAXB = 200005;
const int maxn = 5000005;

char s[MAXS];
int m, na, nb, la[MAXA], lb[MAXB], ra[MAXA], rb[MAXB], w[maxn]; //开始这里 w数组长度开成 MAXA+MAXB了...
struct sub {
	int x, len, typ, id;
	sub(){}
	sub(int x, int len, int typ, int id):x(x), len(len), typ(typ), id(id){}
	inline bool operator <(const sub &o)const {
		return len != o.len ? len > o.len : typ < o.typ;
	}
}Sub[MAXA+MAXB];

namespace _SA { //后缀数组
	int f[MAXS][18], Lg[MAXS], height[MAXS], sa[MAXS], rk[MAXS], c[MAXS], x[MAXS], y[MAXS];
	inline void Pre(int n) { Lg[0] = -1; for(int i = 1; i <= n; ++i) Lg[i] = Lg[i>>1] + 1; }
	inline void Get_Height(int n) {
		for(int i = 1, j, k = 0; i <= n; ++i) if(rk[i] > 1) {
			j = sa[rk[i]-1], k = k ? k-1 : 0;
			while(i+k <= n && j+k <= n && s[i+k] == s[j+k]) ++k;
			height[rk[i]] = k;
		}
	}
	inline void Init_ST(int n) {
		for(int i = 1; i <= n; ++i) f[i][0] = height[i];
		for(int j = 1; j <= Lg[n]; ++j)
			for(int i = 1; i <= n-(1<<j)+1; ++i)
				f[i][j] = min(f[i][j-1], f[i+(1<<(j-1))][j-1]);
	}
	inline void Get_SA(int n, int m) {
		for(int i = 1; i <= m; ++i) c[i] = 0;
		for(int i = 1; i <= n; ++i) ++c[x[i]=s[i]-'a'+1];
		for(int i = 2; i <= m; ++i) c[i] += c[i-1];
		for(int i = n; i >= 1; --i) sa[c[x[i]]--] = i;
		for(int k = 1; k <= n; k<<=1) {
			int p = 0;
			for(int i = n-k+1; i <= n; ++i) y[++p] = i;
			for(int i = 1; i <= n; ++i) if(sa[i] > k) y[++p] = sa[i]-k;
			for(int i = 1; i <= m; ++i) c[i] = 0;
			for(int i = 1; i <= n; ++i) ++c[x[i]];
			for(int i = 2; i <= m; ++i) c[i] += c[i-1];
			for(int i = n; i >= 1; --i) sa[c[x[y[i]]]--] = y[i];
			swap(x, y);
			x[sa[1]] = p = 1;
			for(int i = 2; i <= n; ++i)
				x[sa[i]] = (y[sa[i]] == y[sa[i-1]] && y[sa[i]+k] == y[sa[i-1]+k]) ? p : ++p;
			if((m=p) == n) break;
		}
		for(int i = 1; i <= n; ++i) rk[sa[i]] = i;
		Get_Height(n); Init_ST(n);
	}
}
using namespace _SA;

namespace Gragh { //建图
	int deg[maxn];
	vector<int> G[maxn];
	inline void add(int x, int y) { ++deg[y]; G[x].push_back(y); }
	LL dp[maxn]; int q[maxn];
	inline LL Get_Ans(int N) {
		int l = 0, r = 0;
		for(int i = 1; i <= N; dp[i] = w[i], ++i)
			if(!deg[i]) q[r++] = i;
		LL re = 0;
		while(l < r) { //topo排序
			int u = q[l++];
			re = max(re, dp[u]);
			for(int i = 0, siz = G[u].size(), v; i < siz; ++i) {
				v = G[u][i];
				dp[v] = max(dp[v], dp[u] + w[v]);
				if(!--deg[v]) q[r++] = v;
			}
		}
		if(r < N) re = -1;
		return re;
	}
}
using namespace Gragh;

namespace Segment_Tree { //主席树
	int ls[maxn], rs[maxn], tot;
	void insert(int &now, int l, int r, int x, int id) {
		++tot, ls[tot] = ls[now], rs[tot] = rs[now];
		if(now) add(tot, now);
		w[now = tot] = 0;
		if(l == r) { add(now, id); return; }
		int mid = (l + r) >> 1;
		if(x <= mid) insert(ls[now], l, mid, x, id), add(now, ls[now]);
		else insert(rs[now], mid+1, r, x, id), add(now, rs[now]);
	}
	void addedge(int now, int l, int r, int x, int y, int id) {
		if(!now || y < l || x > r) return;
		if(x <= l && r <= y) { add(id, now); return; }
		int mid = (l + r) >> 1;
		addedge(ls[now], l, mid, x, y, id);
		addedge(rs[now], mid+1, r, x, y, id);
	}
}
using namespace Segment_Tree;
int main () {
	int T; read(T);
	Pre(200000);
	while(T--) {
		int N = 0;
		while(isalpha(s[++N]=getc()));
		Get_SA(--N, 26);

		read(na);
		for(int i = 1; i <= na; ++i)
			read(la[i]), read(ra[i]), Sub[i] = sub(rk[la[i]], w[i]=ra[i]-la[i]+1, 0, i);
		read(nb);
		for(int i = 1; i <= nb; ++i)
			read(lb[i]), read(rb[i]), Sub[na + i] = sub(rk[lb[i]], rb[i]-lb[i]+1, 1, i), w[na+i] = 0;

		sort(Sub + 1, Sub + na + nb + 1);

		tot = na + nb; int rt = 0;
		for(int i = 1; i <= na + nb; ++i) {
			if(!Sub[i].typ) insert(rt, 1, N, Sub[i].x, Sub[i].id);
			else {
				int l = Sub[i].x, r = Sub[i].x;
				for(int j = Lg[l-1]; ~j; --j)
					if(l > (1<<j) && f[l-(1<<j)+1][j] >= Sub[i].len) l -= 1<<j;
				for(int j = Lg[N-r]; ~j; --j)
					if(r < N && f[r+1][j] >= Sub[i].len) r += 1<<j;
				addedge(rt, 1, N, l, r, na + Sub[i].id);
			}
		}

		read(m);
		for(int i = 1, x, y; i <= m; ++i)
			read(x), read(y), add(x, na + y);

		printf("%lld\n", Get_Ans(tot));

		for(int i = 1; i <= tot; ++i)
			deg[i] = ls[i] = rs[i] = 0, G[i].clear();
		ls[0] = rs[0] = 0;
	}
}

常数巨大